JEE Main 2022 · 27 Jun · Shift 1 · Chemistry Q79

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ChemistrySingle correct+4 / -1Question 79 of 90

L-isomer of a compound 'A' ( $\mathrm{C_4H_8O_4}$ ) gives a positive test with $\mathrm{[Ag(NH_3)_2]^+}$ . Treatment of 'A' with acetic anhydride yields triacetate derivative. Compound 'A' produces an optically active compound (B) and an optically inactive compound (C) on treatment with bromine water and $\mathrm{HNO_3}$ respectively. Compound (A) is :

Fischer projection: CHO / HO-C-H / HO-C-H / CH2OH (L-erythrose)

Fischer projection of an aldotetrose: CHO at top, OH on the left at C-2 and C-3, CH2OH at bottom (L-erythrose).

Fischer projection: CHO / H-C-OH / HO-C-H / CH2OH (L-threose)

Fischer projection: OH on the right at C-2 and on the left at C-3 (L-threose).

Fischer projection: CHO / H-C-OH / H-C-OH / CH2OH (D-erythrose)

Fischer projection: OH on the right at C-2 and C-3 (D-erythrose).

Fischer projection: CHO / HO-C-H / H-C-OH / CH2OH (D-threose)

Fischer projection: OH on the left at C-2 and on the right at C-3 (D-threose).

Answer and worked solution

Official answer: A

Clues. $\mathrm{C_4H_8O_4}$ reducing with three OH groups: an aldotetrose. Nitric acid gives an optically inactive (meso) tartaric acid, so A is erythrose (both OH on the same side).

L-isomer. Both OH on the left: structure (A).

(A)

Topics:Carbohydrates Optical isomerism and chirality

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JEE Main 2022 · 27 Jun · Shift 1, Chemistry question 79